Question: Let $g(x)=\dfrac{1}{x^3}$. $g'(3)=$
The strategy We can first rewrite $g(x)$ as a negative power of $x$. Then, the derivative of $g$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have $g'(x)$, we can plug $x=3$ into it to find $g'(3)$. Rewriting the fraction as a negative power $g(x)=\dfrac{1}{x^3}=x^{-3}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(x^{-3}\right) \\\\ &=-3x^{-3-1} \gray{\text{The power rule}} \\\\ &=-3x^{-4} \end{aligned}$ Evaluating $g'(x)$ So we found that $g'(x)=-3x^{-4}$, which can also be written as $-\dfrac{3}{x^4}$. Now let's plug ${x=3}$ : $\begin{aligned} -\dfrac{3}{({3})^4}&=-\dfrac{3}{81} \\\\ &=-\dfrac{1}{27} \end{aligned}$ In conclusion, $g'(3)=-\dfrac{1}{27}$.